The horizontal distance travelled is
{(x=vcos60t),(x=vcostheta(t-0.5)):}
Therefore,
60*1/2*t=60costheta(t-0.5)
30t=60(t-0.5)costheta
t=2tcostheta-costheta
t(2costheta-1)=costheta
t=costheta/(2costheta-1)
The vertical distance travelled is
{(y=vsin60*t-1/2g t^2=52t-4.9t^2),(y=vsintheta(t-0.5)-1/2g(t-0.5)^2=60tsintheta-30sintheta-4.9(t-0.5)^2):}
{(y=52t-4.9t^2),(y=60tsintheta-30sintheta-4.9t^2+4.9t-1.23):}
Therefore,
52t-4.9t^2=60tsintheta-30sintheta-4.9t^2+4.9t-1.23
60tsintheta-30sintheta+4.9t-52t-1.23=0
t(60sintheta-47.1)=30sintheta+1.23
t=(30sintheta+1.23)/(60sintheta-47.1)
The t's must be equal
(30sintheta+1.23)/(60sintheta-47.1)=costheta/(2costheta-1)
(30sintheta+1.23)(2costheta-1)=costheta(60sintheta-47.1)
60sinthetacostheta-30sintheta+2.46costheta-1.23=60sintheta costheta-47.1costheta
-30sintheta+49.56costheta=1.23
49.56costheta-30sintheta=1.23
Solving for theta
rsin(alpha-theta)=rsinalphacostheta-rsinthetacosalpha
{(rcosalpha=-30),(rsinalpha=49.56):}
<=>, r^2=sqrt(30^2+49.56^2)=57.93
{(cosalpha=-30/57.93=-0.558),(sinalpha=49.56/57.93=0.83):}
alpha=58.8^@
Therefore,
57.93sin(58.8-theta)=1.23
sin(58.8-theta)=1.23/57.93=0.021
58.8-theta=1.22
theta=58.8-1.22=57.58^@
And
t=cos(57.6)/(2cos(57.6)-1)=7.42s
x=60cos(57.58)(7.42-0.5)=222.6m
y=52*7.42-4.9*7.42^2=116.1m